Re: Solid of rev and Surface Area

Re: Solid of rev and Surface Area

Okay, taking the origin at the center of the base, the side is a line with slope -5/L passing through (2L, 0). That has equation y= (-5/L)(x- 2L). When x= 3L/2, y= (-5/L)(3L/2- 2L)= (-5/L)(-3L/2)= 15/2. Rotating around the y- axis, you have disks with radius x, thickness dy, so area . That's why you need to integrate . If y= (-5/L)(x- 2L), what is x as a function of y? Now, when you say "surface area" do you mean the slant area only or do you want to incude the area of the top and bottom?

Re: Solid of rev and Surface Area

Oh, blast! That's right. Then you can do this: a line with slope m is of the form y= mx+ b. If we increase x by dx then we increase (decrease for negative m) y by y+ dy= m(x+ dx)+ b subtracting y from the left side and mx+ b from the right we get dy= m dx (which is just saying, again, that for a straight line the slope is the same as the derivative). A right triangle with legs dx and dy has hypotenus of length, by the Pythagorean theorem, . Since the circumference of a circle, of radius r, is , the surface area the product of those two lengths, . That's the formula you had before, right?